Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}+\sqrt{x}\right)-\left(x+1\right)\)
\(=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(\sqrt{x}-1\right)\)
\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)
\(=\left(1+\sqrt{x}\right)^2-\left(2\sqrt{\sqrt{x}}\right)^2\)
\(=\left(1+\sqrt{x}+2\sqrt{\sqrt{x}}\right)\left(1+\sqrt{x}-2\sqrt{\sqrt{x}}\right)\)
\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)
\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)
\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
\(\text{f)}x-2\sqrt{x-1}-a^2\)
\(=\left(\sqrt{x-2}\right)^2\left(\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2}\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}+a\right)\left(\sqrt{x-2\sqrt{x-1}}-a\right)\)
\(ab+b\sqrt{a}+\sqrt{a}+1\)
(đk: \(a\ge0\))
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}+1=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
ĐK: \(x,y\ge0\)
\(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=x\left(\sqrt{x}+\sqrt{y}\right)-y\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{x}-\sqrt{y}\right)\)
a,\(x\sqrt{y}+y\sqrt{x}=\sqrt{x}\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right).\)
c,\(\sqrt{a}-a^2=\sqrt{a}.\left(1-a\sqrt{a}\right)\)
d,\(x-5\sqrt{x}+6=x-3\sqrt{x}-2\sqrt{x}+6\)
\(=\sqrt{x}.\left(\sqrt{x}-3\right)-2.\left(\sqrt{x}-3\right)\)\(=\left(\sqrt{x}-3\right).\left(\sqrt{x}-2\right)\)
a, \(5+\sqrt{5}=\sqrt{5}\left(\sqrt{5}+1\right)\)
b, \(a-2\sqrt{a}=\sqrt{a}\left(\sqrt{a}-2\right)\)
c, \(x-\sqrt{xy}=\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\)
d, \(x-y-\sqrt{x}-\sqrt{y}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-1\right)\)
a) 2a−4b=2(a−2b)2a−4b=2(a−2b)
c) 2ax−2ay+2a=2a(x−y+1)2ax−2ay+2a=2a(x−y+1)
e) 3xy(x−4)−9x(4−x)=3x(x−4)(y+3)3xy(x−4)−9x(4−x)=3x(x−4)(y+3)
b,d xem lại đề
không hiểu
what are you doing?